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(a) What are the frequency and wavelength of a photon emitted during transition from n = 5 state to n = 2 state in the hydrogen atom ? (b) In which region of the electromagnetic spectrum will this radiation lie ?

Answer» <html><body><p></p>Solution :(a) According to Rhdberg formula, `bar(v) = <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` <br/> Here, `R = 109, 677 cm^(-1), n_(2) = 5, n_(1) = 2` <br/> `:. bar(v) 109,677 ((1)/(2^(2)) - (1)/(5^(2))) cm^(-1) = 109,677 xx (21)/(100) cm^(-1) = 23032.2 cm^(-1)` <br/> `lamda = (1)/(v) = (1)/(23032.2 cm^(-1)) = 434 xx 10^(-7) cm= 434 xx 10^(-9) m = 434 <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>` <br/> `v = (c)/(lamda) = (3 xx 10^(8) ms^(-1))/(434 xx 10^(-9)m) = 6.91 xx 10^(15) s^(-1)` <br/> (b) The wavelength, as calculated above, lies in the visible region. Otherwise too, as the jump is on the <a href="https://interviewquestions.tuteehub.com/tag/2nd-300843" style="font-weight:bold;" target="_blank" title="Click to know more about 2ND">2ND</a> orbit, the line will belong to Balmer series and hence lie in the visible region. <br/> Alternatively, as discussed later under <a href="https://interviewquestions.tuteehub.com/tag/bohr-400140" style="font-weight:bold;" target="_blank" title="Click to know more about BOHR">BOHR</a>'s <a href="https://interviewquestions.tuteehub.com/tag/model-942" style="font-weight:bold;" target="_blank" title="Click to know more about MODEL">MODEL</a>, on page 2//31, energy of an electron in the nth shell is given by `E_(n) = (-21.8 xx 10^(-19))/(n^(2))J` <br/> `:.` Energy released when electron undergoes transition from n = 5 to n = 2, will be <br/> `Delta E = E_(5) - E_(2) = (-21.8 xx 10^(-19))/(5^(2)) - (- (21.8 xx 10^(19))/(2^(2))) J = -21.8 xx 10^(-19) ((1)/(25) - (1)/(4)) J` <br/> `= 21.8 xx 10^(-19) xx (21)/(100) J = 4.58 xx 10^(-19)J` <br/> But`Delta E = hv " " :. v = (Delta E)/(h)= (4.58 xx 10^(-19)J)/(6.626 xx 10^(-34)Js ) = 6.91 xx 10^(14) s^(-1) or Hz` <br/> `lamda = (c)/(v) = (3.0 xx 10^(8) ms^(-1))/(6.91 xx 10^(14) s^(-1)) = 434 xx 10^(-9) m = 434nm`</body></html>


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